已知y=x^5 sinx,则y^n=
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y'=(sinx)'+(cos^5x)'=cosx-5cos^4x*sinx符合函数的求导,其实就是对式子一层一层的求导.对于cos^5x,可以看做a=cosx,b=a^5.再一层层求导就可以了
因为已知x,β是锐角,sinx=4/5所以cosx=3/5又因为cos(x+β)=5/13所以sin(x+β)=12/13cos(x+β)=cosxcosβ-sinxsinβ=3/5cosβ-4/5s
这类题重点在于转换y=cos2x+(sinx)^2-cosx=(cosx)^2-(sinx)^2+(sinx)^2-cosx=(cosx)^2-cosx=(cosx-1/2)^2-1/41.当cosx
/>sinx+siny=1/3,cosx-cosy=1/5两式分别平方得sin²x+sin²y+2sinxsiny=1/9cos²x+cos²y-2cosxco
y=3x/5原式=x/(x+3x/5)+(3x/5)/[x-3x/5]-(9x^3/25)/(x^3-9x^3/25)=8/3-3/2-9/16=29/48
∵sin(x+y)=1,∴x+y=π2+2kπ,k∈Z,∴y=-x+π2+2kπ,∴sin(2y+x)=sin(-2x+π+4kπ+x)=sin(π-x)=-sinx=-13,故答案为:-13.
可以采取对数求导由y=(sinx)^x得lny=ln(sinx)^x=xln(sinx)两边求导得到1/y*y'=ln(sinx)+x*cosx*1/sinx所以得到y'=(sinx)^x*ln(si
y=sin²x+sinx+cosx+2=(1-cos2x)/2+√2sin(x+л/4)+2=(1/2)*sin(2x+л/2)+√2*sin(x+л/4)+5/2;=(1/2)*sin(2
y=(sinx+cosx)^2+2cos^x=2+sin2x+cos2x=2+√2sin(2x+π/4)ymax=2+√2,ymin=2-√2.2kπ+π/2≤2x+π/4≤2kπ+3π/22kπ+π
(-x^2*sinx-2x*cosx+2sinx)/(x^3)再问:可以具体一点儿吗再答:(sinx/x)'=(x*cosx-sinx)/(x^2)(sinx/x)''=[(cosx-x*sinx-c
tan(x-y)=(tanx-tany)/(1+tanx*tany)=[(sinxcosy-cosxsinx)/(cosxcosy)/[(cosxcosy+sinxsiny)/(cosxcosy)]=
y=x^sinx两边取对数lny=ln(x^sinx)=sinx*lnx然后两边对x求导(注意y是关于x的函数,所以lny其实是一个复函数)(1/y)*y'=cosx*lnx+sinx/x即y'/y=
y=x/sinx+sinx/xy'=(sinx-xcosx)/sin²x+(xcosx-sinx)/x²
因为,-π/2
∵x、y都是锐角,sinx=4/5,cos(x+y)=5/13∴cosx=3/5,sin(x+y)=12/13∵cos(x+y)=cosx*cosy-sinx*sinysin(x+y)=sinx*co
cos(x-y)cosx+sin(x-y)sinx=cos(x-y-x)=cos(-y)=cosy=3/5
(1)y=(x-2)/x+1=1-2/x+1=2-2/xy'=2/x²(2)y=tanxy'=sec²x(3)y=x^2lnx+sinxy'=2xlnx+x²/x+cos
sinx*cosy=1sinx=cosy=1或sinx=cosy=-1cosx=siny=0因此cos(x+y)=cosxcosy-sinxsiny=0
y'=f'(x+sinx)(1+cosx)y''=f''(x+sinx)(1+cosx)^2+f'(x+sinx)(1-1/1+x^2)=f"(x+sinx)(1+cosx)^2+f'(x+sinx)
y'=根号下sinx比x